#1928

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Cost to Reach Destination in Time

Minimize the travel cost in a graph while adhering to a time constraint using state transition dynamic programming.

Problem Statement

You are in a country with n cities, numbered from 0 to n - 1, connected by bidirectional roads. Each road is represented by a 2D array, edges, where edges[i] = [xi, yi, timei] represents a road between cities xi and yi that takes timei minutes to travel. Some cities may have multiple roads between them, but there are no roads connecting a city to itself.

You are tasked with starting at city 0 and reaching city n - 1 within a given maximum travel time, maxTime. Along your journey, you must pay passing fees for each city you traverse, represented by an array passingFees where passingFees[j] is the fee for passing through city j. Your goal is to minimize the total fee while reaching the destination in the allowed time.

Examples

Example 1

Input: maxTime = 30, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3]

Output: 11

The path to take is 0 -> 1 -> 2 -> 5, which takes 30 minutes and has $11 worth of passing fees.

Example 2

Input: maxTime = 29, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3]

Output: 48

The path to take is 0 -> 3 -> 4 -> 5, which takes 26 minutes and has $48 worth of passing fees. You cannot take path 0 -> 1 -> 2 -> 5 since it would take too long.

Example 3

Input: maxTime = 25, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3]

Output: -1

There is no way to reach city 5 from city 0 within 25 minutes.

Constraints

1 <= maxTime <= 1000
n == passingFees.length
2 <= n <= 1000
n - 1 <= edges.length <= 1000
0 <= xi, yi <= n - 1
1 <= timei <= 1000
1 <= passingFees[j] <= 1000
The graph may contain multiple edges between two nodes.
The graph does not contain self loops.

Solution Approach

State Transition Dynamic Programming

Model the problem using dynamic programming where each state represents a city and the time taken to reach that city. Create a new graph with each node representing a city at a specific time. Use a DP array to track the minimum cost of reaching each city at each time. Transition between cities based on the available edges while considering both time and cost.

Graph Representation and Time Handling

Represent the cities and roads as a graph and handle the time constraint by tracking the journey’s duration at each city. At each step, make sure that traveling to the next city doesn't exceed the time limit, adjusting your cost calculation accordingly.

Priority Queue Optimization

To efficiently manage the states, use a priority queue to explore cities in the order of the minimum cost, prioritizing paths with the least passing fee. This approach ensures that you always explore the least costly path while maintaining the time constraint.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the chosen approach. With dynamic programming and priority queues, the time complexity is typically O(n * maxTime * log n), where n is the number of cities. Space complexity can be O(n * maxTime), as we store the cost for each city and time combination.

What Interviewers Usually Probe

Look for an efficient way to manage state transitions across cities and times.
Pay attention to how the candidate handles edge cases, such as when it's impossible to reach the destination in time.
Evaluate the candidate's approach to optimizing performance using priority queues or dynamic programming.

Common Pitfalls or Variants

Common pitfalls

Ignoring the time constraint and exploring paths that exceed maxTime.
Not considering multiple paths between two cities with different travel times.
Overlooking edge cases where the destination is unreachable within the time limit.

Follow-up variants

Introduce multiple source and destination cities.
Modify the problem by allowing one-way roads.
Vary the passing fees by making them dynamic based on time of day or other conditions.

FAQ

How does state transition dynamic programming help in solving Minimum Cost to Reach Destination in Time?

State transition dynamic programming allows you to model the problem where each state represents a city at a specific time, efficiently managing both time and cost during the journey.

What is the time complexity of solving the Minimum Cost to Reach Destination in Time?

The time complexity is typically O(n * maxTime * log n), where n is the number of cities and maxTime is the time limit.

How can I optimize my solution for the Minimum Cost to Reach Destination in Time?

Using priority queues to explore cities in the order of the minimum cost is a great optimization strategy, ensuring the least costly paths are explored first.

What happens if no path exists to reach the destination within the given time?

If no valid path exists within the time limit, the output should be -1, indicating that it is impossible to reach the destination in time.

Can this problem be solved without dynamic programming?

While dynamic programming is the most efficient approach, it is possible to solve the problem with other methods such as depth-first search or breadth-first search, though they may be less optimal in handling time constraints and costs.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the minimum cost to reach city $j$ from city $0$ after $i$ minutes. Initially, $f[0][0] = \textit{passingFees}[0]$, and the rest $f[0][j] = +\infty$.

class Solution:
    def minCost(
        self, maxTime: int, edges: List[List[int]], passingFees: List[int]
    ) -> int:
        m, n = maxTime, len(passingFees)
        f = [[inf] * n for _ in range(m + 1)]
        f[0][0] = passingFees[0]
        for i in range(1, m + 1):
            for x, y, t in edges:
                if t <= i:
                    f[i][x] = min(f[i][x], f[i - t][y] + passingFees[x])
                    f[i][y] = min(f[i][y], f[i - t][x] + passingFees[y])
        ans = min(f[i][n - 1] for i in range(m + 1))
        return ans if ans < inf else -1

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